I've taken some probability classes before, but it's been a while. I'm looking for (1) some resource -- website, paper, software, book -- etc on conditional probabilities. Something which accessible to a junior/senior level undergrad. Specifically, I'm interested in the relationships described by the following example:

Suppose that the nodes $A$ and $B$ are independent, and the node $C$ depends on both (I understand this may be an example of what's called a directed acyclic graph, if that's important).
The type of dependencies associated with the nodes I described above rely on truth tables. For example, $A$ and $B$'s probability distributions:

$$ D(A) = \begin{array}{c|c} \text{True} & \text{False} \\ \hline 0.3 & 0.7 \end{array}, \quad D(B) = \begin{array}{c|c} \text{True} & \text{False} \\ \hline 0.01 & 0.99 \end{array} $$

And $C$'s (in percentages)

enter image description here

Then, without observable evidence, the probability that $C$ is FALSE* is $ \approx 0.9688$.
How does one arrive at that value?
Furthermore, how would I generalize this behavior to a larger graph, with possibly more complicated stuff like $A$ and $B$ not being independent.

  • $\begingroup$ I´m not sure about the topic here. But I´ve calculated that the probability C is true is $\approx 1-0.9688=0.0312$ $\endgroup$ – callculus Sep 28 '18 at 14:00
  • $\begingroup$ Have you checked your solution? $\endgroup$ – callculus Sep 28 '18 at 14:20
  • $\begingroup$ Yep, sorry about that, I meant $P(C)$ is false. So you're right. $\endgroup$ – Zduff Sep 28 '18 at 14:30
  • $\begingroup$ I have no recommendation for liturature. I can only show how to calculate $P(\overline C)$. $\endgroup$ – callculus Sep 28 '18 at 14:33
  • $\begingroup$ Yep, that's all I need, if you don't mind. $\endgroup$ – Zduff Sep 28 '18 at 14:39

To calculate the probability that $C$ is false we calculate all combinations of $A$ und $B$ which lead to $\overline C$.

$\overline C$ is the complementary event of $C$.

$$P(\overline C)=P(A)\cdot P(B|A)\cdot P(\overline C|A\cap B)+P(\overline A)\cdot P(B|\overline A)\cdot P(\overline C|\overline A\cap B)+P( A)\cdot P(\overline B| A)\cdot P(\overline C|A\cap \overline B)+P(\overline A)\cdot P(\overline B|\overline A)\cdot P(\overline C|\overline A\cap \overline B)$$

The events are: $A:=$Event A is true, $B:=$Event B is true,$\overline A:=$Event A is false, $\overline B:=$Event B is false,$C:=$Event C is true, $\overline C:=$Event C is false

$A$ and $B$ are independent. Therefore $P(B|A)=P(B)$. This is the same for all combinations with the complementary events

$$P(\overline C)=P(A)\cdot P(B)\cdot P(\overline C|A\cap B)+P(\overline A)\cdot P(B)\cdot P(\overline C|\overline A\cap B)+P( A)\cdot P(\overline B)\cdot P(\overline C|A\cap \overline B)+P(\overline A)\cdot P(\overline B)\cdot P(\overline C|\overline A\cap \overline B)$$

And the conditional probability $P(\overline C|A\cap B)=100 \%=1$ is given in the table. This is the last row and third column. Similar for the combinations of the events A, B. Therefore

$$P(\overline C)=0.7\cdot 0.99\cdot 0.999+0.99\cdot 0.3\cdot 0.9+0.01\cdot 0.7\cdot 0.88+0.3\cdot 0.01\cdot 1 \approx 0.9688$$

And $P(C)=1-P(\overline C)$. This result can be also obtained by using the equivalent formula which is used to calculate $P(\overline C)$.


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